Is there a metric space such that for a sequence of nonempty subsets $K_1 \supset K_2 ....$ the intersection of all subsets is empty?

Question

Is there a metric space $(X,d)$ such that for a sequence of $\emptyset \neq K_j \subset X$ with $K_1 \supset K_2 ....$ such that the intersection of all $\bigcap _{j\in \mathbb{N}}K_j$ is empty?

It seems very obvious that since all $K_j$ are not empty, there is an element in $K_1$ that is also in all $K_i$ with $i \geq 1$ and that this element this is an element of the intersection, which thus is nonempty.

However in the task it was specified that the metric space is compact and the subsets $K_j$ are closed. I have used neither in the argument above and cannot figure out a counterexample where it would be necessary to have the conditions to the sets.

Answer 1

Both conditions are necessary for the theorem to hold. Here are two examples:

Sets are not closed:

Consider the metric space $[0,1]$ with the usual metric, and the sets $K_n = (0,\frac{1}{n}]$. This space is compact, and the sets $K_n$ are nonempty and $K_n \supset K_{n+1}$. However, the intersection of these sets is empty: remark that the sets $K_n$ are not closed.

Space is not compact:

Consider the metric space $(0,1]$ with the usual metric, and the sets $K_n = (0,\frac{1}{n}]$. These sets are nonempty, we have $K_n \supset K_{n+1}$, and importantly: the sets are closed in $(0,1]$ (you may have to convince yourself of this). However, again, the intersection of these sets is empty: remark that the space is not compact.

Answer 2

Sure, the intersection can be empty: let $X =\mathbb{R}$ in the standard metric and $K_n = [n, \infty)$ which are all closed. And if $x \in \mathbb{R}$, let $N$ be some integer above $x$, then $x \notin K_N$ so $x \notin \bigcap_n K_n$, so that $\bigcap_n K_n =\emptyset$.

The nice thing is that this cannot happen if $X$ or some $K_n$ is compact.

Answer 3

Yes. Take the reals (with the usual topology) and $K_n=[n,+\infty)$.